--- title: "Bode plot: 10 S(S+3)/((S+1)(S²+2S+16))" subject: "Electrical Circuit Theory" chapter: 5 key: "semester_3/electrical_circuit_theory:bode-10s-s3-s1-s2-2s" source: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-10s-s3-s1-s2-2s site: BCT Engineering --- # Bode plot: 10 S(S+3)/((S+1)(S²+2S+16)) Electrical Circuit Theory · Chapter 5: Frequency Response of Network - Asked in: 2067 Asar (old course), 8 marks - Video: [1. Bode Plot Solved Example: Steps, Procedure, Parameters, and Stability](https://www.youtube.com/watch?v=lTnqZVNReF8) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_circuit_theory/bode-10s-s3-s1-s2-2s - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_circuit_theory/chapter-5 ## Question Sketch Bode-plot for the transfer function given by G(S) = 10 S(S + 3) / [(S + 1)(S² + 2S + 16)]. ## Short answer Express G(s) as 1.875 s(1+s/3)/[(1+s)(1+0.125s+s²/16)]. Corner freqs: 1,3,4 rad/s. Magnitude: initial slope +20 dB/dec, 5.5 dB at ω=1, changes: −20 at 1, +20 at 3, −40 at 4. Quadratic pole (ωn=4, ζ=0.25) adds ~6 dB peak. Phase: 90° to −90°. ## Answer A Bode plot has two curves against log ω: the gain 20 log|H(jω)| in dB and the phase ∠H(jω). The asymptotic (straight-line) magnitude plot is drawn by writing H(s) in time-constant form, finding the corner frequencies and changing the slope by ±20 dB/decade for each simple factor (±40 for each quadratic factor). #### Step 1: Standard form ```text G(s) = 10 s (s + 3) / [ (s + 1)(s² + 2s + 16) ] ``` ```text H(s) = 1.875 × s (1 + s/3) / [ (1 + s/1) (1 + 0.125 s + s²/16) ] ``` Constant term K₀ = 1.875, so 20 log K₀ = 5.46 dB. #### Step 2: Corner frequencies | Corner ω (rad/s) | Factor | Slope change | |---|---|---| | 1 | pole 1/(1 + s/1) | −20 dB/dec | | 3 | zero (1 + s/3) | +20 dB/dec | | 4 | quadratic pole, ωn = 4, ζ = 0.25 | −40 dB/dec | The quadratic s² + 2s + 16 has ωn = 4 rad/s and 2ζωn = 2, so ζ = 0.25. There is a zero at the origin, so the low-frequency slope is +20 dB/decade. #### Step 3: Magnitude plot The net number of zeros at the origin is 1, so the low-frequency asymptote has slope +20 dB/decade and passes through 5.5 dB at ω = 1 rad/s. The slope then changes at each corner as below. | ω (rad/s) | Slope after corner | Asymptotic gain (dB) | Phase (°) | |---|---|---|---| | 0.1 | +20 | −14.5 | 85 | | 1 | 0 | 5.5 | 56 | | 3 | +20 | 5.5 | 23 | | 4 | −20 | 8 | −23 | Asymptotic magnitude plot (each * is the asymptote; 1 row = 5 dB): ```text 10 | * | ******** *** 0 | *** ** | *** −10 | *** |** ++-----------+-----------+ 0.1 1 10 ω (rad/s) ``` At ωn = 4 the quadratic pole has ζ = 0.25, so the true curve rises about 6 dB above the asymptote there (resonant peak, −20 log 2ζ); the actual peak is near ωn. #### Step 4: Phase plot Low-frequency phase = 90° × (zeros at origin − poles at origin) = 90°. Each simple factor changes the phase by 90° (zero: +, pole: −), spread over the range 0.1ωc to 10ωc with 45° at ωc; each quadratic factor changes it by 180°, with ±90° at ωn. High-frequency phase = 90° × (m − n) = −90°. Exact phase values at the corners are in the table above; join them with a smooth curve starting at 90° and ending at −90°. Accuracy check: at the first corner ω = 1 rad/s the exact gain is 3.4 dB, the asymptote gives 5.5 dB. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the starting slope and phase? A. +20 dB/dec, +90 deg B. -20 dB/dec, -90 deg C. 0 dB/dec, 0 deg D. +20 dB/dec, -90 deg Answer: A. One zero at the origin gives +20 dB/dec and +90 deg. (Key point: Low-frequency slope) 2. What is K0? A. 10 (20 dB) B. 1.875 (5.46 dB) C. 30 D. 0.625 Answer: B. K0 = 10 x 3 /(1 x 16) = 1.875. (Key point: Constant term) 3. For s^2 + 2s + 16, what are omega_n, zeta and the peak? A. omega_n = 16, zeta = 0.125, about 12 dB B. omega_n = 4, zeta = 0.5, about 0 dB C. omega_n = 4, zeta = 0.25, about 6 dB peak D. omega_n = 2, zeta = 0.25, about 6 dB Answer: C. 2 zeta x 4 = 2 so zeta = 0.25, and -20 log 0.5 = 6 dB. (Key point: Quadratic) 4. What is the high-frequency phase? A. -270 deg B. +90 deg C. -180 deg D. -90 deg Answer: D. 90 x (m - n) = 90 x (2 - 3) = -90 deg. (Key point: High-frequency phase) ## Close variants set in other exams - [Bode plot: 64(S+2)/(S(S²+0.5S)(S²+3.2S+64))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-64s2-s-s2-05s-s2) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 200(S+1)/(S(S+5)(S²+2S+100))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-200s1-s-s5-s2-2s) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 20(S+1)/((S²+4S+2)(S²+5S))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-20s1-s2-4s-2-s2) — Electrical Circuit Theory, chapter 5, asked 4× ## Related in Electrical Circuit Theory - [Bode plot: 40(s+1)/((2s²+10s)(s²+2s+10))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-40s1-2s2-10s-s2-2s) — Electrical Circuit Theory, chapter 5 - [Bode plot: 20(s+1)/(s(s+5)(s²+2s+10))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-20s1-s-s5-s2-2s) — Electrical Circuit Theory, chapter 5 - [Bode plot: 2s²(S+5)/((S²+22S+40)(S+10))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-2s2-s5-s2-22s-40) — Electrical Circuit Theory, chapter 5, asked 2× - [Bode plot: 20(S+5)/(S(S²+2S+10)(S²+21S+20))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-20s5-s-s2-2s-10) — Electrical Circuit Theory, chapter 5, asked 2× - [Bode plot: 10(S+10)/(S(S²+5S+4)(S+40))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-10s10-s-s2-5s-4) — Electrical Circuit Theory, chapter 5, asked 4× - [Bode plot: 20(s+5)/(S(S+20)(s²+80s+200))](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Abode-20s5-s-s20-s2-80s) — Electrical Circuit Theory, chapter 5, asked 4× ## Related in other subjects - [Transfer function from asymptotic Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-from-bode-asymptote) — Control Systems, chapter 6, asked 2× - [Determining transfer function from Bode plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-to-tf) — Control Systems, chapter 6, asked 2× - [Transfer function from Bode plot with example](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abode-to-tf-example) — Control Systems, chapter 6, asked 2× - [DT transfer function; drawing a Bode plot](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-transfer-function-bode) — Signals and Systems, chapter 6